The Jordan multiplication semigroup of matrix algebras is the full endomorphism semigroup

TL;DR

This paper proves that for matrix algebras over fields with characteristic not 2, the Jordan multiplication semigroup generated by all multiplication operators equals the entire endomorphism semigroup, showing a comprehensive algebraic structure.

Contribution

It establishes that every linear endomorphism of the underlying vector space can be expressed as a composition of Jordan multiplication operators, a novel linear-algebraic result.

Findings

The Jordan multiplication semigroup equals the full endomorphism semigroup.

Every linear endomorphism is a composition of multiplication operators.

The proof involves constructing elementary transvections and analyzing determinant surjectivity.

Abstract

Let $\mathbb{K}$ be a field of characteristic different from $2$, and let $M_n(\mathbb{K})$ be the algebra of all $n\times n$ matrices over $\mathbb{K}$. We consider the corresponding special Jordan algebra $\mathcal{A}:=M_n(\mathbb{K})^+$ with symmetrized product $A\circ B:=(AB+BA)/2$, and write $\mathcal{A}_{\mathrm v}:=M_n(\mathbb{K})$ for the underlying $\mathbb{K}$-vector space of $\mathcal{A}$. For $A\in\mathcal{A}$, let $\mathrm{L}_A(X):=A\circ X$ be the multiplication operator. We consider the Jordan multiplication semigroup generated by all multiplication operators, \[ \mathrm{JMS}(\mathcal{A}):=\langle \mathrm{L}_A:A\in\mathcal{A}\rangle\subseteq \mathrm{End}_{\mathbb{K}}(\mathcal{A}_{\mathrm v}). \] We prove that $\mathrm{JMS}(\mathcal{A})=\mathrm{End}_{\mathbb{K}}(\mathcal{A}_{\mathrm v})$. Equivalently, every $\mathbb{K}$-linear endomorphism of $\mathcal{A}_{\mathrm v}$ is a composition of multiplication operators. The proof is primarily linear-algebraic. The main step is to show that $\mathrm{SL}(\mathcal{A}_{\mathrm v})\subseteq \mathrm{JMS}(\mathcal{A})$ by constructing elementary transvections inside the semigroup. We then prove determinant surjectivity on the unit group of $\mathrm{JMS}(\mathcal{A})$ and combine it with the existence of a singular element of rank $n^2-1$ to obtain the full endomorphism semigroup. In the finite-field case, the determinant-surjectivity step is established via Jacobi-sum estimates.